Velocity at Window Top for Dropped Ball – Rankers Physics
Topic: Kinematics
Subtopic: Motion Under Gravity

Velocity at Window Top for Dropped Ball

A person sitting in the ground floor of a building notices through the window, of height \(1.5\text{ m}\), a ball dropped from the roof of the building crosses the window in \(0.1\text{ s}\). What is the velocity of the ball when it is at the topmost point of the window? \((g = 10\text{ m/s}^2)\)

(2020-Covid)

14.5 m/s
4.5 m/s
20 m/s
15.5 m/s

Solution:

Concept: Equations of motion under gravity for a specific interval.
Formula: \(h = ut + \frac{1}{2}gt^2\).
Solution: Let \(u\) be velocity at window top. Given \(h=1.5\text{ m}\,\text{ }t=0.1\text{ s}\,\text{ }g=10\text{ m/s}^2\). \(1.5 = u(0.1) + \frac{1}{2}(10)(0.1)^2\). \(1.5 = 0.1u + 0.05\). \(1.45 = 0.1u\) => \(u = 14.5\text{ m/s}\).

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